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12-1
Characteristics of F-Distribution
 There
is a “family” of F Distributions.
Each
member of the family is determined by
two parameters: the numerator degrees of
freedom and the denominator degrees of freedom.
F cannot be negative, and it is a continuous
distribution.
The F distribution is positively skewed.
Its values range from 0 to  . As F   the
curve approaches the X-axis.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-2
Test for Equal Variances
For
the two tail test, the test statistic is given
by:
F 
s12
s 22
2
1s
and s 22 are the sample variances for
the two samples.
The null hypothesis is rejected if the
computed value of the test statistic is greater
than the critical value.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-3
EXAMPLE 1
 Colin,
a stockbroker at Critical Securities,
reported that the mean rate of return on a sample
of 10 internet stocks was 12.6 percent with a
standard deviation of 3.9 percent. The mean rate
of return on a sample of 8 utility stocks was 10.9
percent with a standard deviation of 3.5 percent.
At the .05 significance level, can Colin conclude
that there is more variation in the software
stocks?
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-4
EXAMPLE 1
Step
continued
1: The hypotheses are:
H 0 :  I2   U2
H1 :  I2   U2
Step
2: The significance level is .05.
Step
3: The test statistic is the F
distribution.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-5
Example 1
continued
Step
4: H0 is rejected if F>3.68. The degrees of
freedom are 9 in the numerator and 7 in
the denominator.
Step 5: The value of F is computed as
follows.
2
F
(3.9)
(3.5)
2
 1.2416
H0 is not rejected. There is insufficient evidence
to show more variation in the internet stocks.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-6
Underlying Assumptions for ANOVA
 The
F distribution is also used for testing
whether two or more sample means came from
the same or equal populations.
This technique is called analysis of
variance or ANOVA.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-7
Analysis of Variance Procedure
 The
Null Hypothesis is that the population means
are the same.
The
Alternative Hypothesis is that at least one of
the means is different.
The Test Statistic is the F distribution.
The Decision rule is to reject the null hypothesis
if F (computed) is greater than F (table) with
numerator and denominator degrees of freedom.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-8
Analysis of Variance Procedure
 If
there are k populations being sampled, the
numerator degrees of freedom is k – 1.
 If there are a total of n observations the
denominator degrees of freedom is n – k.
 The test statistic is computed by:
SST k  1
F
SSE n  k 
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-9
Analysis of Variance Procedure
SS
Total is the total sum of squares.
2
(

X
)
2
SS total  X 
n
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Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-10
Analysis of Variance Procedure
 SST
is the treatment sum of squares.
T
SST  
 n
 c
2
c
 X 



n

2
TC is the column total, nc is the number of
observations in each column, X the sum of all the
observations, and n the total number of
observations.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-11
Analysis of Variance Procedure
SSE
is the sum of squares error.
SSE  SS total - SST
McGraw-Hill/Irwin
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12-12
EXAMPLE 2
 Rosenbaum
Restaurants specialize in meals for
senior citizens. Katy Polsby, President, recently
developed a new meat loaf dinner. Before
making it a part of the regular menu she decides
to test it in several of her restaurants. She would
like to know if there is a difference in the mean
number of dinners sold per day at the Anyor,
Loris, and Lander restaurants. Use the .05
significance level.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-13
Example 2
Aynor
13
12
14
12
Tc
nc
51
4
McGraw-Hill/Irwin
continued
Loris
10
12
13
11
46
4
Lander
18
16
17
17
17
85
5
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-14
Example 2 continued
The
SS total is:
2
2
(

X
)


182
SS total  X 2 
 2634 
 86
n
13
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Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-15
Example 2 continued
The
SST is:
 Tc2
SST  
 nc

 X 2


n

 512 46 2 85 2



 4
4
5

 76 .25
McGraw-Hill/Irwin
 (182 ) 2


13

Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-16
Example 2 continued
The
SSE is:
SSE = SS total – SST = 86 – 76.25 = 9.75
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12-17
EXAMPLE 2
Step
Step
continued
1: H0: 1 = 2 = 3
H1: Treatment means are not the
same
2: H0 is rejected if F>4.10. There are 2
df in the numerator and 10 df in the
denominator.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-18
Example 2 continued
To
find the value of F:
SST k  1
F
SSE n  k 
76 3  1

 39 .10
9.75 13  3
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Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-19
Example 2
continued
The
decision is to reject the null hypothesis.
 The treatment means are not the same.
The mean number of meals sold at the three
locations is not the same.
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12-20
Inferences About Treatment Means
When
we reject the null hypothesis that the
means are equal, we may want to know which
treatment means differ.
One
of the simplest procedures is
through the use of confidence intervals.
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Confidence Interval for the
Difference Between Two Means
12-21
 1 1
 X1  X 2   t MSE  n  n 
1
2
where
t is obtained from the t table with degrees
of freedom (n - k).
MSE = [SSE/(n - k)]
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
12-22
EXAMPLE 3
 From
EXAMPLE 2 develop a 95% confidence interval
for the difference in the mean number of meat loaf
dinners sold in Lander and Aynor. Can Katy conclude
that there is a difference between the two restaurants?
 1 1
(17  12.75)  2.228 .975  
 4 5
4.25  148
.  (2.77,5.73)
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12-23
Example 3continued
Because
zero is not in the interval, we conclude
that this pair of means differs.
The
mean number of meals sold in Aynor is
different from Lander.
McGraw-Hill/Irwin
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.